1
GATE CSE 2010
MCQ (Single Correct Answer)
+2
-0.6
A computer system has an $$L1$$ cache, an $$L2$$ cache, and a main memory unit connected as shown below. The block size in $$L1$$ cache is $$4$$ words. The block size in $$L2$$ cache is $$16$$ words. The memory access times are $$2$$ nanoseconds, $$20$$ nanoseconds and $$200$$ nanoseconds for $$L1$$ cache, $$L2$$ cache and main memory unit respectively. GATE CSE 2010 Computer Organization - Memory Interfacing Question 22 English

When there is a miss in $$L1$$ cache and a hit in $$L2$$ cache, a block is transferred from $$L2$$ cache to $$L1$$ cache. What is the time taken for this transfer?

A
$$2$$ nanoseconds
B
$$20$$ nanoseconds
C
$$22$$ nanoseconds
D
$$88$$ nanoseconds
2
GATE CSE 2010
MCQ (Single Correct Answer)
+2
-0.6
A $$5$$-stage pipelined processor has Instruction Fetch $$(IF),$$ Instruction Decode $$(ID),$$ Operand Fetch $$(OF),$$ Perform Operation $$(PO)$$ and Write Operand $$(WO)$$ stages. The $$IF, ID, OF$$ and $$WO$$ stages take $$1$$ clock cycle each for any instruction. The $$PO$$ stage takes $$1$$ clock cycle for $$ADD$$ and $$SUB$$ instructions, $$3$$ clock cycles for $$MUL$$ instruction, and $$6$$ clock cycles for $$DIV$$ instruction respectively. Operand forwarding is used in the pipeline. What is the number of clock cycles needed to execute the following sequence of instructions? GATE CSE 2010 Computer Organization - Pipelining Question 25 English
A
$$13$$
B
$$15$$
C
$$17$$
D
$$19$$
3
GATE CSE 2010
MCQ (Single Correct Answer)
+2
-0.6
A hash table of length 10 uses open addressing with hash function h(k)=k mod 10, and linear probing. After inserting 6 values into an empty hash table, the table is as shown below GATE CSE 2010 Data Structures - Hashing Question 9 English How many different insertion sequences of the key values using the same hash function and linear probing will result in the hash table shown above?
A
10
B
20
C
30
D
40
4
GATE CSE 2010
MCQ (Single Correct Answer)
+2
-0.6
The following C function takes a simply-linked list as input argument. It modifies the list by moving the last element to the front of the list and returns the modified list. Some part of the code is left blank.
typedef struct node { 
    int value; 
    struct node *next; 
} Node;
 
Node *move_to_front(Node *head) { 
     Node *p, *q; 
     if ((head = = NULL: || (head->next = = NULL)) return head; 
     q = NULL; p = head; 
     while (p-> next !=NULL) { 
       q=P; 
       p=p->next; 
     } 
     return head; 
} 
Choose the correct alternative to replace the blank line.
A
q = NULL; p->next = head; head = p;
B
q->next = NULL; head = p; p->next = head;
C
head = p; p->next = q; q->next = NULL;
D
q->next = NULL; p->next = head; head = p;
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